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| 1. |
The angle of a quadrilateral are in AP whose common difference is 10 find the angles |
| Answer» Let the required angles be (a - 3d)°, (a - d)\xa0°, (a + d)\xa0°\xa0and (a + 3d)\xa0°Common difference = (a - d) - (a- 3d) = a - d - a + 3d = 2d\xa0We are given that Common difference = 10°{tex}\\therefore{/tex}2d = 10°\xa0= d = 5°We know that sum of four angles of quadrilateral = 360°{tex}\\Rightarrow{/tex} (a-3d)o +(a-d)o+(a+d)o+(a+3d)o=360o4a=360oa={tex}\\frac{{360}}{4}{/tex}=90o{tex}\\therefore{/tex}a=90o and d=5oFirst angle = (a - 3d)°\xa0= (90 - 3\xa0×\xa05)\xa0°\xa0= 75°Second angle = (a - d)° = (90 - 5)\xa0°\xa0= 85°Third angle = (a + d)° = (90 + 5°) = 95°Fourth angle = (a + 3d)° = (90 + 3\xa0×\xa05)° = 105° | |